Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $k = \dfrac{q + 1}{q^2 + 3q - 4} \times \dfrac{q + 4}{3q + 3} $
Answer: First factor the quadratic. $k = \dfrac{q + 1}{(q + 4)(q - 1)} \times \dfrac{q + 4}{3q + 3} $ Then factor out any other terms. $k = \dfrac{q + 1}{(q + 4)(q - 1)} \times \dfrac{q + 4}{3(q + 1)} $ Then multiply the two numerators and multiply the two denominators. $k = \dfrac{ (q + 1) \times (q + 4) } { (q + 4)(q - 1) \times 3(q + 1) } $ $k = \dfrac{ (q + 1)(q + 4)}{ 3(q + 4)(q - 1)(q + 1)} $ Notice that $(q + 1)$ and $(q + 4)$ appear in both the numerator and denominator so we can cancel them. $k = \dfrac{ (q + 1)\cancel{(q + 4)}}{ 3\cancel{(q + 4)}(q - 1)(q + 1)} $ We are dividing by $q + 4$ , so $q + 4 \neq 0$ Therefore, $q \neq -4$ $k = \dfrac{ \cancel{(q + 1)}\cancel{(q + 4)}}{ 3\cancel{(q + 4)}(q - 1)\cancel{(q + 1)}} $ We are dividing by $q + 1$ , so $q + 1 \neq 0$ Therefore, $q \neq -1$ $k = \dfrac{1}{3(q - 1)} ; \space q \neq -4 ; \space q \neq -1 $